Nonhomogeneous Systems — Question 10

Question 10 – Solution

Strategy. A rotating frame converts the forcing into accumulated scalar amplitude; cancelling its total integral removes the persistent free rotation.

Step 1: Integrate in rotating coordinates. Since $R'=JR$ and $R^{-1}=R(-t)$, writing $X=R(t)Z$ gives $Z'=e^{-t^2}(1,0)^T$. Define $G(t)=\int_0^t e^{-s^2}\,ds$. Then $$\boxed{X(t)=R(t)\left[c+G(t)\binom 10\right].}$$ For $c=0$ the state is $G(t)(\cos t,\sin t)^T$. Differentiation verifies the equation, including the non-elementary amplitude derivative.

Step 2: Interpret the zero-state asymptotics. Let $L=\sqrt\pi/2$. For $c=0$, the norm is $G(t)\uparrow L$, strictly increasing at every finite time, so the solution is not periodic. At times $2\pi n$ and $2\pi n+\pi$ it tends to $(L,0)^T$ and $(-L,0)^T$, so no point limit exists. The input decays, but the undamped homogeneous rotation retains the accumulated amplitude; there is no dissipative mechanism to erase it.

Step 3: Select the unique cancelling initial state. Because rotations preserve length, $\|X(t)\|\to\|c+L(1,0)^T\|$. Thus $X(t)\to 0$ exactly when $\boxed{c=(-L,0)^T}$. For that state, $$\boxed{X(t)=-R(t)\binom 10\int_t^\infty e^{-s^2}\,ds.}$$ Any other initial state leaves a nonzero limiting amplitude.

Step 4: Bound the remaining tail. For $s\ge t>0$, $1\le s/t$. Therefore $$0\le\int_t^\infty e^{-s^2}\,ds \le\frac 1t\int_t^\infty s e^{-s^2}\,ds =\frac{e^{-t^2}}{2t}.$$ The selected solution consequently obeys $\boxed{\|X(t)\|\le e^{-t^2}/(2t)\quad(t>0)}$. This is a tail estimate, not a formula at $t=0$; the initial norm there is $L$. The definite integral and the bound avoid inventing an elementary antiderivative or replacing the exact response by numerical samples.